令b=π/4cos(a+π/4)=cosa·cosπ/4-sin·asinπ/4=(cosa-sina)(·√2/2)sin(a-π/4)=sina cosπ/4-cos·s inπ/4=(sina·-cosa·)(√2/2)cos(π/4+a)=sin(a-π/4),(cosa-sina)(·√2/2)=(sina·-cosa·)(√2/2)cosa...這是兩角的和差公式嗎？還沒學(xué)呢。。是兩角的和差公式cos(α+β)=cosα·cosβ-sinα·sinβ cos(α-β)=cosα·cosβ+sinα·sinβ sin(α±β)=sinα·cosβ±cosα·sinβtan(α+β)=(tanα+tanβ)/(1-tanα·tanβ) tan(α-β)=(tanα-tanβ)/(1+tanα·tanβ)還有沒有其他的方法？我還沒學(xué)兩角的和差公式。令b=π/4∵cos(π/4+a)=sin(a-π/4),∴cos(π/4+a)=sin(π/2-(a+π/4))=sin(π/2-a-π/4)=sin(π/4-a)=sin(a-π/4),∴π/4-a=a-π/4a=π/4∵π/2-a=π/2-π/4=π/4∴f(a)=f(π/2-a)=sin(a+π/4)+cos(a-π/4)=sin(π/4+π/4)+cos(π/4-π/4) =1+1=2前面算的有錯(cuò)=(sina+cosa)(√2/2)+(cosa+sina)( √2/2)